# What is int (-x^3-2x-3 ) / (7x^4+ 5 x -1 )?

Mar 1, 2016

$- .65697 \ln | x - .19785 | - .01191 \ln | x + .95254 | - .51407 \ln | \frac{.78462}{\sqrt{{x}^{2} - .75468 x + .75801}} | - .05036 {\tan}^{- 1} \left(\frac{x - .37734}{.78462}\right) + c o n s t .$

#### Explanation:

What are the roots of the polynomial
$7 {x}^{4} + 5 x - 1$?

This is a depressed quartic function and I recommend this 2 sources (the first is to show work, the second for a fast solution) to resolve it:

http://www.sosmath.com/algebra/factor/fac12/fac12.html

http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php

From the root finder we get:

${x}_{1} = .19785$
${x}_{2} = - .95254$
${x}_{3} = .37734 + i .78462$
${x}_{4} = .37734 - i .78462$

For not to deal with complex numbers we have
$\left(x - {x}_{3}\right) \left(x - {x}_{4}\right) = {x}^{2} - .75468 x + .75801$

The original function can be rewritten in partial fractions in this way
$\frac{- {x}^{3} - 2 x - 3}{7 {x}^{4} + 5 x - 1} = \frac{A}{x - .19785} + \frac{B}{x + .95254} + \frac{C x + D}{{x}^{2} - .75468 x + .75801}$

For $x = 0 , - 1 , 1 \mathmr{and} 2$ we get

$\frac{A}{-} .19785 + \frac{B}{.95254} + 0 \cdot C + \frac{D}{.75468} = 3$
$\frac{A}{-} 1.19785 + \frac{B}{.04746} + \frac{- C + D}{5.51269} = 0$
$\frac{A}{.80215} + \frac{B}{1.95254} + \frac{C + D}{1.00333} = - \frac{6}{11}$
$\frac{A}{1.80295} + \frac{B}{2.95254} + \frac{2 C + D}{3.24865} = - \frac{15}{121}$

Or

$\left[\begin{matrix}- 5.05433 & 1.04982 & 0 & 1.31924 \\ - .83483 & 21.07038 & - .39798 & .39798 \\ 1.24665 & .51215 & .99668 & .99668 \\ .55489 & .33869 & .61564 & .30782\end{matrix}\right] \left[\begin{matrix}A \\ B \\ C \\ D\end{matrix}\right] = \left[\begin{matrix}3 \\ 0 \\ - \frac{6}{11} \\ - \frac{15}{121}\end{matrix}\right]$

Solving this system of variables we get

$A = - .65697$
$B = - .01191$
$C = .51407$
$D = - .23349$

So the original expression becomes

$- .65697 \int \frac{\mathrm{dx}}{x - .19785} - .01191 \int \frac{\mathrm{dx}}{x + .95254} + \int \frac{.51407 x - .23349}{{x}^{2} - .75468 x + .75801} \mathrm{dx}$ [ $\alpha$ ]

Let's resolve the last part of the expression, the only one that poses a challenge
$\int \frac{.51407 x - .23349}{{x}^{2} - .75468 x + .75801} \mathrm{dx} =$

${\left(x - .37734\right)}^{2} = {x}^{2} - .75468 + .14239$
=> ${x}^{2} - .75468 x + .75801 = {\left(x - .37734\right)}^{2} + .61562$
$\left(x - .37734\right) = \sqrt{.61562} \tan y = .78462 \tan y$
$\mathrm{dx} = .78462 {\sec}^{2} y \mathrm{dy}$
How many units of $\left(x - .37734\right)$ are there in the numerator?
$\frac{.51407 x - .23349}{x - .37734} = .51407 - \frac{.03951}{x - .37734}$

That's why now we are dealing with
=.51407int (.78462tany*.78462cancel(sec^2y))/(.61562cancel(sec^2y))dy-.03951int (.78462cancel(sec^2y))/(.61562cancel(sec^2y)dy
$= .51407 \int \tan y \mathrm{dy} - .05036 \int \mathrm{dy}$
$= - .51407 \ln | \cos y | - .05036 y$

$\tan y = \frac{x - .37734}{.78462}$ => $\sin y = \frac{x - .37734}{.78462} \cos y$
=>${\sin}^{2} y + {\cos}^{2} y = 1$ => $\left(\frac{{x}^{2} - .75468 x + .14239}{.61562} + 1\right) {\cos}^{2} y = 1$ => $\cos y = \frac{.78462}{\sqrt{{x}^{2} - .75468 x + .75801}}$
$\to = - .51407 \ln | \frac{.78462}{\sqrt{{x}^{2} - .75468 x + .75801}} | - .05036 {\tan}^{- 1} \left(\frac{x - .37734}{.78462}\right)$

Therefore expression [ $\alpha$ ] becomes
$- .65697 \ln | x - .19785 | - .01191 \ln | x + .95254 | - .51407 \ln | \frac{.78462}{\sqrt{{x}^{2} - .75468 x + .75801}} | - .05036 {\tan}^{- 1} \left(\frac{x - .37734}{.78462}\right) + c o n s t .$